We show that for $y_{1},\ldots,y_{k} \in \mathbb{Z}$ it follows that there exists a y in $\mathbb{Z}$ so that $y\equiv y_{1} \pmod {m_{1}}, y\equiv y_{k} \pmod {m_{k}} \Leftrightarrow m_{1}|y-y_{1},\ldots, m_{k}| y- y_k$